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Pls Help - Calc. HW dy/dx problem

Pls Help Calc HW Dydx Problem class=

Sagot :

Space

Answer:

[tex]\displaystyle y' = \frac{-2}{x \ln (10)[\log (x) - 2]^2}[/tex]

General Formulas and Concepts:

Calculus

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Addition/Subtraction]:                                                         [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]

Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Quotient Rule]:                                                                           [tex]\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}[/tex]

Step-by-step explanation:

Step 1: Define

Identify.

[tex]\displaystyle y = \frac{\log (x)}{\log (x) - 2}[/tex]

Step 2: Differentiate

  1. [Function] Derivative Rule [Quotient Rule]:                                                 [tex]\displaystyle y' = \frac{[\log (x) - 2][\log (x)]' - [\log (x) - 2]'[\log (x)]}{[\log (x) - 2]^2}[/tex]
  2. Rewrite [Derivative Rule - Addition/Subtraction]:                                       [tex]\displaystyle y' = \frac{[\log (x) - 2][\log (x)]' - [\log (x)' - 2'][\log (x)]}{[\log (x) - 2]^2}[/tex]
  3. Logarithmic Differentiation:                                                                         [tex]\displaystyle y' = \frac{[\log (x) - 2]\frac{1}{\ln (10)x} - [\frac{1}{\ln (10)x} - 2'][\log (x)]}{[\log (x) - 2]^2}[/tex]
  4. Derivative Rule [Basic Power Rule]:                                                             [tex]\displaystyle y' = \frac{[\log (x) - 2]\frac{1}{\ln (10)x} - \frac{1}{\ln (10)x}[\log (x)]}{[\log (x) - 2]^2}[/tex]
  5. Simplify:                                                                                                         [tex]\displaystyle y' = \frac{\frac{\log (x) - 2}{\ln (10)x} - \frac{\log (x)}{\ln (10)x}}{[\log (x) - 2]^2}[/tex]
  6. Simplify:                                                                                                         [tex]\displaystyle y' = \frac{\frac{-2}{\ln (10)x}}{[\log (x) - 2]^2}[/tex]
  7. Rewrite:                                                                                                         [tex]\displaystyle y' = \frac{-2}{x \ln (10)[\log (x) - 2]^2}[/tex]

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

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